Question

Can you perceive the shift in frequency produced when you pull a tuning fork toward you at\({\rm{10}}{\rm{.0}}\;{\rm{m/s}}\)on a day when the speed of sound is \({\rm{344}}\;{\rm{m/s}}\)? To answer this question, calculate the factor by which the frequency shifts and see if it is greater than\({\rm{0}}{\rm{.300\% }}\).

Short answer

The frequency shift is \(2.99\% \).

Step-by-step solution

Given Data

The speed of sound is \(v = 344\;{\rm{m/s}}\).

The speed of the source is \({v_s} = 10\;{\rm{m/s}}\)

The speed of the observer is zero.

Doppler shift

The Doppler shift is the shift in actual frequency and the apparent frequency due to relative motion between the source and the listener.

Calculation of the Doppler Shift

The apparent frequency from Doppler Effect is,

\(\begin{array}{c}f' = f\left( {\frac{{v - {v_o}}}{{v - {v_s}}}} \right)\\\frac{{f'}}{f} = \left( {\frac{{v - {v_o}}}{{v - {v_s}}}} \right)\end{array}\)

Now, plugging the values,

\(\begin{array}{c}\frac{{f'}}{f} = \left( {\frac{{344 - 0}}{{344 - 10}}} \right)\\\frac{{f'}}{f} = \frac{{344}}{{334}}\\\frac{{f'}}{f} = 1.0299\end{array}\)

The shift is,

\(\begin{array}{c}\frac{{f' - f}}{f} \times 100\% \\ = \left( {\frac{{f'}}{f} - 1} \right) \times 100\% \\ = \left( {1.0299 - 1} \right) \times 100\% \\ = 2.99\% \end{array}\)